Determine all positive integers $n$ such that $n={d(n)}^2$.
Problem
Source:
Tags: Divisor Functions
Peter
30.08.2007 15:58
Obviously so $ n$ is odd (a square has an odd number of divisors). Let $ n=p_{1}^{2k_{1}}\cdots p_{m}^{2k_{m}}$, then we have to solve \[ (2k_{1}+1)\cdots (2k_{m}+1)=p_{1}^{k_{1}}\cdots p_{m}^{k_{m}},\] and since $ 2x+1<3^{x}$ for all $ x>1$ we have $ p_{1}\le 3$ and $ k_{1}\le 1$, while all other primes cannot appear. So $ n\in\{1,9\}$.
Wave-Particle
11.10.2015 20:45
Note that $n$ is a perfect square. So we know that $n=(p_1^{e_1}\cdot p_2^{e_2}...p_k^{e_k})^2$. We know that $\sqrt{n}=p_1^{e_1}\cdot p_2^{e_2}...p_k^{e_k}=d(n)=(2e_1+1)(2e_2+1)...(2e_k+1)$ so it follows that since $3^x>2x+1$ for $x>1$ that if $k_1>1$ we cannot have $p_1\ge3$. If $k_1=1$ then we have $p_1=3$ which is a solution. So $n=9$ is one solution. If $k_1>1$ and $p_1=2$ this won't work because we know that $d(n)$ is odd implying that $n$ is odd. The only case we need to check now is $n=1$ which is indeed a solution. So $n=1,9$ are our solutions.